Set, Relations and FunctionsPYQ Jun 23Question 1947 of 217
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If R\displaystyle R be a relation defined on the set of Natural numbers as "x\displaystyle x R\displaystyle R y⇔(x−y)\displaystyle y \Leftrightarrow (x-y) is divisible by 5\displaystyle 5" ∀x,y∈N\displaystyle \forall x, y \in N then the relation R\displaystyle R is

Options

AEquivalence
BAnti-symmetric
CSymmetric but not transitive
DSymmetric but not reflexive
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Correct Answer

✅ Option a — Equivalence

All Options:

  • AEquivalence
  • BAnti-symmetric
  • CSymmetric but not transitive
  • DSymmetric but not reflexive

Detailed Solution & Explanation

Let R\displaystyle R be defined on the set of natural numbers N\displaystyle \mathbb{N} as:
xRy  ⟺  (x−y) is divisible by 5x R y \iff (x - y) \text{ is divisible by } 5
Let's check the three properties of an equivalence relation:
1. **Reflexive**: For any x∈N\displaystyle x \in \mathbb{N}:
x−x=0x - x = 0
Since 0\displaystyle 0 is divisible by 5 (0=5×0\displaystyle 0 = 5 \times 0), xRx\displaystyle x R x holds. The relation is reflexive.
2. **Symmetric**: If xRy\displaystyle x R y, then (x−y)\displaystyle (x - y) is divisible by 5, i.e., x−y=5k\displaystyle x - y = 5k for some integer k\displaystyle k.
y−x=−(x−y)=−5k=5(−k)y - x = -(x - y) = -5k = 5(-k)
Since −k\displaystyle -k is an integer, (y−x)\displaystyle (y - x) is also divisible by 5, meaning yRx\displaystyle y R x. The relation is symmetric.
3. **Transitive**: If xRy\displaystyle x R y and yRz\displaystyle y R z, then x−y=5k1\displaystyle x - y = 5k_1 and y−z=5k2\displaystyle y - z = 5k_2 for some integers k1\displaystyle k_1 and k2\displaystyle k_2. Adding these:
(x−y)+(y−z)=5k1+5k2(x - y) + (y - z) = 5k_1 + 5k_2
x−z=5(k1+k2)x - z = 5(k_1 + k_2)
Since k1+k2\displaystyle k_1 + k_2 is an integer, (x−z)\displaystyle (x - z) is divisible by 5, meaning xRz\displaystyle x R z. The relation is transitive.
Since R\displaystyle R is reflexive, symmetric, and transitive, it is an **Equivalence** relation.
Hence, **Option A** is the correct answer.

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